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Chemical Kinetics

Question

In the Arrhenius equation for certain reaction, the values of the frequency factor and energy of activation are 4 x 10¹³ sec^–1 and 98.6 kJ/mol respectively. If the reaction is of first order, at what temperature will its half-life be 10 minutes.

Solution

We have given that

A = 4 \times 10^{13} s^{- 1} E_{a} = 98.6 kJ / mol = 98600 J / mol t_{1 / 2} = 10 \min .

∴

k = \frac{0.693}{10} = 0.0693 \min^{- 1} = \frac{0.0693}{60} k = 0.001155 s^{- 1}

We know that

\log k = \log A - \frac{E_{a}}{2.303 R^{+ 1}}

\frac{E_{a}}{2.303 R} . \frac{1}{T} = logA - logB = \log 4 \times 10^{13} - \log 0.00155 = 13.602 + 2.937 \frac{E_{a}}{2.303 R} . \frac{1}{T} = 16.539

T = \frac{98600}{2.303 \times 8.314 \times 16.539} = 311.4 K